Calculating Power Factor for 3-Phase AC System of 100kW @ 12500V

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The discussion focuses on calculating the required capacitor size to improve the power factor of a balanced 3-phase AC system from 0.75 to 0.90 lagging, with an initial load of 100kW at 12500V. The calculations indicate that the size of the capacitor needed per phase is approximately 0.416mF, assuming a frequency of 60Hz. The real power remains unchanged at 520kW, as the resistance component of the impedance is not altered. The total reactive power before the capacitor is 321kVAR, which reduces to 151.62kVAR after the capacitor is added. Overall, the approach to calculating the power factor improvement and capacitor size appears to be correct.
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The system with a balanced 3 phase load of 100kW at a 0.75 power factor, the system voltage is 12500V line to line, grounded wye.

What size of capacitor can bring the power factor to 0.90 lagging, then what will be the real power?
 
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This was my work.
Would anyone please verify my approach is correct?

Initial 3 phase load of Ptotal =520kW, Pf1=0.85, and Vline=31540volts
3*Pf=Ptotal
Pf=520kW/3=173.33kW

For grounded wye,
Vf=Vline/SQRT3=18209volts

Pf1=0.85,
Pf1=Sf1 * (cosf1), where Sf1= Pf1/(cosf1)
Sf1=173.33W/0.85=203.92kVA
Qf1=SQRT(Sf12-Pf12)=107kVAR


Pf2=0.96,
Sf2=Pf2/(cosf2) = 173.33kW/0.96=180.55kVA
Qf2=SQRT(Sf22-Pf22)=50.54kVAR

Xf(reactive) = Vf2/Qf=182092/Qf=331.567E6/ Qf
Xf(before) = 331.567E6/107k=3086Ω
Qfcap= Qf1-Qf1=107k-50.54k=56460VAR
Xfcap= Vfcap2/Qfcap= 331.567E6/56460= 5872 Ω

a) The size of the capacitor per phase C is defined as
C=1/(2*p*f* Xfcap)=1/(376.99*5872)=0.416mF since it is assumed f=60Hz

b) Real power should remain same since we are not changing anything about the resistance part of the impedance so
Ptotal = 3* Pf =520kW
Pf=173.33kW per phase

C) Qtotal = 3*Qf
So Qtotal before = 3*107kVAR=321kVAR and Qtotal after = 3*50.54kVAR=151.62kVAR
 

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